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2 years ago

Jonah runs a bakery, and his most popular item is cinnamon raisin bread. He has two options to purchase raisins.

Mathematics

2 answers:

Effectus [21]2 years ago

7 0

Answer:

lvvies can you please help me again

Step-by-step explanation:

garri49 [273]2 years ago

3 0

Answer:

C. $0.19/ounce

Step-by-step explanation:

We know that 16 ounces = 1 pound but we have 5 pounds,

So, we do 5 times 16.

We get 80 (ounces) and we divided ounces by total cost ($15.25)

We get our answer:

$0.19/ounce

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Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

3 years ago

How many runners ran fewer than 4 km?

Vinvika [58]

10 runners ran fewer than 4 km

4 0

3 years ago

Read 2 more answers

A researcher wishes to be 95% confident that her estimate of the true proportion of individuals who travel overseas is within 3%

andre [41]

Answer: a) 683 b) 1067

Step-by-step explanation:

The confidence interval for population proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

a) Given : Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}}=\pm1.96$

Margin of error : $E=0.03$

Formula to calculate the sample size needed for interval estimate of population proportion :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.2(0.8)(\dfrac{1.96}{0.03})^2=682.951111111\approx683$

Hence, the required sample size would be 683 .

b) If no estimate of the sample proportion is available then the formula to calculate sample size will be :-

$n=0.25(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.25(\dfrac{1.96}{0.03})^2=1067.11111111\approx1067$

Hence, the required sample size would be 1067 .

3 0

3 years ago

What is the Range of this Graph? ASAP

Flura [38]

Answer:

The range would be {-4,-3,1,6}

Step-by-step explanation:

Individual points are bracketed

6 0

3 years ago

Jose buys a drink and dinner at a local restaurant. The dinner was $9.89 and the drink was $1.99. Jose would like to leave a 15%

My name is Ann [436]

I believe it is 1.78

4 0

3 years ago

Read 2 more answers